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Mk nhầm nha câu đầu chỉ có 1 cái x-1 + x -2 thôi ko có cái đằng sau nhé ! giá trị tuyệt đối thì vẫn giữ nguyên !
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
a) (2x-1)(x+2)=0
<=> 2x-1=0
Hoặc x+2=0
<=>2x=1
Hoặc x=-2
<=>x=1/2
Hoặcx=-2
Xl nha mk ko bt làm mấy câu kia
a) \(2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^3-2x=0\)
\(\Leftrightarrow x\left(x^2-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)
c) \(x^6+1=0\)
\(\Leftrightarrow x^6=-1\)
Ta có : \(x^6\ge0\) với mọi x
Mà : -1 < 0
=> Vô nghiệm
d) \(x^3+2x=0\)
\(\Leftrightarrow x\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=-2\left(loại\right)\end{matrix}\right.\)
e) \(x^5+8x^2=0\)
\(\Leftrightarrow x^2\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
f) \(x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)
g) \(\left(x+\dfrac{1}{2}\right)\left(x^2-\dfrac{4}{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x^2-\dfrac{4}{5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x^2=\dfrac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\sqrt{\dfrac{4}{5}}\end{matrix}\right.\)
\(TH1:a,2\left|x-3\right|+\left|2x+5\right|=11\)
\(\Rightarrow2x-6+2x+5=11\)
\(\Rightarrow4x-1=11\)
\(\Rightarrow4x=12\)
\(\Rightarrow x=3\)
\(TH2:2\left|x-3\right|+\left|2x+5\right|=11\)
\(\Rightarrow-2x+6-2x-5=11\)
\(\Rightarrow-4x+1=11\)
\(\Rightarrow-4x=10\)
\(\Rightarrow x=-2,5\)
\(TH1:b,\left|x-3\right|+\left|5-x\right|+2\left|x-4\right|=2.2\)
\(\Rightarrow x-3+5-x+2x-8=4\)
\(\Rightarrow2x-6=4\)
\(\Rightarrow x=5\)
\(TH2:\left|x-3\right|+\left|5-x\right|+2\left|x-4\right|=4\)
\(\Rightarrow-x+3-5+x-2x+8=4\)
\(\Rightarrow-2x+6=4\)
\(\Rightarrow x=1\)
a: TH1: x<1
A=1-x+2-x=3-2x
TH2; 1<=x<2
A=x-1+2-x=1
TH3: x>=2
A=x-1+x-2=2x-3
b: TH1: x<5/2
B=5-2x+3-x+x-2=-2x+6
TH2: 5/2<=x<3
B=2x-5+3-x+x-2=2x-4
TH3: x>=3
B=x-3+2x-5+x-2=4x-10
c: TH1: x<-3/2
C=-2x-3-(5-x)+2x
=-2x-3-5+x+2x
=x-8
TH2: -3/2<=x<5
C=2x+3-(5-x)+2x=4x+3-5+x=5x-2
TH3: x>=5
C=2x+3-(x-5)+2x=4x+3-x+5=3x+8